Image charge distance from the center of a sphere

[Living_Dog]

hi all,

This question is from Griffiths', Intro. to Electrodynamics, Example 3.2, to wit:

"A point charge q is situated a distance 'a' from the center of a grounded conducting sphere of radius 'R' (Fig. 3.12). Find the potential outside the sphere."

Note: a > R. Griffiths' solution does not provide the explanation on how the value of 'b' (the distance of the image charge, q', which is inside the sphere b < R) was obtained. The potential I started with is:

V(r) = k(q'/r_1' + q/r_2),

where r'_1 and r_2 are found using the law of cosines from their respective charges to the observation point outside the sphere and k := 1/4pi*epsilon_o

The image charge, q', can be found using the fact that the sphere is grounded:

q' = - (R - a)/( b - R)

I tried using V = 0 at r = infinity, but I get a wrong answer:

b = 2*R - a, (2 times R)

whereas, the correct result is:

b = R^2/a (R-squared).

Thanks in advance for any help you may give me,
-LD

 

[StatusX]

First thing is check your expression for q'. You're missing a q, and I think its upside down.

 

[siddharth]

hi all,

This question is from Griffiths', Intro. to Electrodynamics, Example 3.2, to wit:

"A point charge q is situated a distance 'a' from the center of a grounded conducting sphere of radius 'R' (Fig. 3.12). Find the potential outside the sphere."

Note: a > R. Griffiths' solution does not provide the explanation on how the value of 'b' (the distance of the image charge, q', which is inside the sphere b < R) was obtained.

But, that's the whole point! Griffiths is using the method of images. Here, he replaces the sphere and the point charge, with two completely different point charges. Why does he do this? Because it works! The potential of this seemingly arbitrary configuration satisfies the boundary conditions(ie, V=0 ar R and [itex]\infty[/itex]). So, by the first uniqueness theorem, this is also the solution to your original question.

 

[Living_Dog]

First thing is check your expression for q'. You're missing a q, and I think its upside down.

yeah, sorry, it was a typo: q' = q (R - b)/(a - R).

-LD

 

[Living_Dog]

But, that's the whole point! Griffiths is using the method of images. Here, he replaces the sphere and the point charge, with two completely different point charges. Why does he do this? Because it works! The potential of this seemingly arbitrary configuration satisfies the boundary conditions(ie, V=0 ar R and [itex]\infty[/itex]). So, by the first uniqueness theorem, this is also the solution to your original question.

I understand the method, I don't know how he got the result for the distance of the image charge, 'b'.

-LD

 

[StatusX]

Ok, well V=0 at infinity automatically, since there is only a finite amount of charge, so you need another meaningful relation, say that the potential is zero at the other end of the sphere. This will allow you to solve for q' and b. Then you only need to verify that this gives you zero potential everywhere on the sphere, which is only a lucky coincidence particular to this geometry.

 

[siddharth]

yeah, sorry, it was a typo: q' = q (R - b)/(a - R).

-LD

How do you get that? That isn't right.

I'll say this again, the important point here, is that the new point charge q' is totally different. It just happens to be -(R/a)q. It's not possible to know beforehand about q' and b.

In fact, Griffiths says, "The first person who solved the problem this way cannot have known in advance what image charge q' to use and where to put it. Presumably, he(she?) started with an arbitrary charge at an arbitrary point inside the sphere, calculated the potential on the sphere, and then discovered that with q' and b just right the potential on the sphere vanishes. But, its really a miracle that any choice does the job."

 

[StatusX]

siddharth, I think what he's trying to do is, knowing in advance that there is some point charge that can be placed inside the sphere that causes the potential to vanish everywhere on the surface, to try to solve for its charge and position. Yes, it is a little contrived, although it does give an idea where those values that Griffiths pulled out of a hat actually came from. And I believe the expression he found for q' will reduce to yours once you solve for b and plug it in.

 

[siddharth]

Oh, I see! I thought the OP had some confusion over the "method of images". Thanks for that StatusX.

 

[Living_Dog]

siddharth, I think what he's trying to do is, knowing in advance that there is some point charge that can be placed inside the sphere that causes the potential to vanish everywhere on the surface, to try to solve for its charge and position. Yes, it is a little contrived, although it does give an idea where those values that Griffiths pulled out of a hat actually came from. And I believe the expression he found for q' will reduce to yours once you solve for b and plug it in.

Exactly, thank you.

Here it is again. Isn't there another (independent) equation which allows 'b' to be determined? I already tried V = 0 at r = [tex]\infty[/tex] but to no avail.

This is my concern - what if I get this on a qualifying exam... how can I show that I found the correct 'b'?? But if in the end the final answer is: "you have to just remember it" (like a trig. identity), then I'm fine with that... I've thought about this so long b = R²/a is burned into my neural pattern!

-LD
_______________________________________
my faith: http://www.angelfire.com/ny5/jbc33

 

[StatusX]

As I said before, try setting the potential to zero at the other end of the sphere, ie, the point antipodal (polar opposite) to the point you already solved for.

 

[Living_Dog]

As I said before, try setting the potential to zero at the other end of the sphere, ie, the point antipodal (polar opposite) to the point you already solved for.

You did it. It fell out as easy as the initial result for q'. Thanks a million!

-LD
________________________________________
my faith: http://www.angelfire.com/ny5/jbc33/